Question: Solve: $\dfrac{3}{4} + \dfrac{1}{6} - \dfrac{2}{3} = $
Explanation: Let's look at the multiples of each denominator and see which multiples they have in common. Denominator Multiples ${4}$ $4, 8, \underline{{12}}$ ${6}$ $6, \underline{{12}}, 18$ ${3}$ $3, 6, 9, \underline{{12}}$ The least common multiple is ${12}$. Let's use multiplication to make each fraction have a denominator of $12$. $\begin{aligned} &{\dfrac{3}{4}}=\dfrac{{3} \times 3}{{4} \times 3} = {\dfrac{9}{12}}\\\\ &{\dfrac{1}{6}}=\dfrac{{1} \times 2}{{6} \times2} = {\dfrac{2}{12}}\\\\ &{\dfrac{2}{3}}=\dfrac{{2} \times 4}{{3} \times4} = {\dfrac{8}{12}} \end{aligned}$ $\begin{aligned} &{\dfrac{3}{4}} + {\dfrac{1}{6}} - {\dfrac{2}{3}}\\\\ =& {\dfrac{9}{12}} + {\dfrac{2}{12}} - {\dfrac{8}{12}}\\\\ =&\dfrac{9 + {2} - {8}}{12}\\\\ =&\dfrac{11 - 8}{12}\\\\ =&\dfrac{3}{12} \end{aligned}$ $\dfrac34 + \dfrac{1}{6} - \dfrac{2}{3} = \dfrac{3}{12}$ $\dfrac3{12}$ can also be written as $\dfrac14$.